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ferlop
modified 3 years ago

Latch made with BJTs

4
14
129
01:42:03
You have to use resistors around 270 Ohm to drive the transistor into heavier conduction to keep the other one off. I bench tested this successfully with 3904s.
published 9 years ago
ferlop
9 years ago
Debugger is to busy doing space time continuum experiments to answer a simple question, correction, he answers when you don't ask???
nikisalli
9 years ago
Nice 👍 i'll try this
thebugger
9 years ago
What's the questionâť“:D
thebugger
9 years ago
Niki don't try this. You'll only burn out two perfectly good LED's and maybe two transistors. Why you ask? Well normally LED's have Vf of 2V and at 5V you'll fry them. Second you're currently providing 15mA of current into the base of the transistors to produce a collector current of 30mA. This gives you a gain of 2 while normally low power modern transistors have a hFe of around 100-1000. You'd usually only need 100uA or so to drive the transistor at sufficient levels. At 15mA you'll damage it. And finally I wouldn't use this topology at all for a bistable latch. I'd use one somewhat different, a couple more resistors but a lot more stability
ferlop
9 years ago
Niki the LEDs are just to show the flipping around, yeah don't use them, as far as burning the transistors, 3904 is good for 500ma so, what is the issue? Like I said debug answers the question when you are not even thinking it.
thebugger
9 years ago
Yes it's called heuristic thinking. 500mA is the collector current I'm talking about the base current. I'd try something more like this http://everycircuit.com/circuit/5365377237778432
ferlop
9 years ago
Debug, you didn't publish it? Ok, so if Niki wants to use base resistors, that is up to him, this topology works fine as a plain bi-stable latch on the board, in real, no smoke.
thebugger
9 years ago
Yeah but it won't last long, at all. Base current of low power transistors is meant to be in the uA range. You always, ALWAYS protect the base from overcurrent in saturation mode circuits.
thebugger
9 years ago
One more problem with your circuit. Imagine the LED's are for 2V and you don't manage to blow them up in the first attempt, due to very low gain transistors or something. Each time you press a switch you'll be exposing the LED to the full voltage of the power supply, and that would surely damage them. Maybe not momentarily, but in time, for sure. That's one of the reasons I put the switches at the base, but your bases are connected to the collector, so you can't do that either
ferlop
9 years ago
I guess I should have pointed out that I don't hold the design rights on that circuit. I just thought it was an interesting circuit.
thebugger
9 years ago
It is, just badly designed. Other than that flip flops are nice.
ferlop
9 years ago
That is why I was asking, when I published Bistable latches, my question to you was: why the bipolar topology would not work? and finally we get the answer, a lot of dancing around along with it, but thanks, I really neglected the base current consideration. You are right! I am going to test your circuit on the bench. Note: If you remove the LEDs from your R circuit, and connecting the 100ohm resistors directly to the collectors, in simulation it does not flip, it just remains latched to one side. It works with 9k, not with 10k.
thebugger
9 years ago
I never saw the question. Yes with 100ohm it wouldn't work because the transistors barely have enough base current to saturate let alone to latch. You need to decrease the value of the base resistors to 5k or so

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